 Namaste, Myself, Mr. Virajdar Bala Saheb, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Solapu. In this video lecture, we will discuss applications to electrical circuit Part 1. Learning Outcome At the end of this session, students will be able to solve problems of RL and RC series electrical circuits with initial condition. Let us start with basic fundamentals and definitions. These are the symbols and their units are required to form a differential equation of a given electrical circuits. Name, Symbol and their units, current is denoted by I measured in ampere, charge is denoted by Q measured in coulomb, time is denoted by T measured in second, resistance is denoted by R measured in ohms, inductance is denoted by L measured in Henry, capacitance is denoted by capital C measured in parades. In order to avoid notations, we have to remember the definition of current and statement of Kirchhoff's voltage law. First of all see the definition of current. Current is defined as rate of change of charge Q on a plate of condenser with respect to time t that is I is equal to dQ by dt. Now statement of Kirchhoff's voltage law is a very important law is required to form a differential equation of a given electrical circuits. It is stated as the algebraic sum of the voltage drop around any closed circuit is equal to the resultant electromotive force in the circuit. In order to above two definition, we required voltage drop across elements of a circuit. First one, voltage drop across the resistance R is equal to R into I, secondly the voltage drop across the inductance L is equal to L into di by dt, third one the voltage drop across the capacitance C is equal to Q by C. With the help of these, we can write the differential equation of Q 1 circuit. Let us see the differential equation of R L circuit. A circuit involving resistance R and inductance L along with an electromotive force denoted by E of t in a series shown in figure. This is the R L circuit with electromotive force E of t in a series. Where V sub pixel indicates voltage drop across inductance L which is equal to L di by dt. V R indicates voltage drop across resistance which is equal to R into I. Now by Kirchhoff's voltage law that is sum of voltage drop across L and R is equal to electromotive force applied to the circuit. Therefore, L into di by dt plus R into I is equal to E of t. This is the differential equation of R L circuit in a series. This is the first order first degree ordinary differential equation belongs to linear form here I dependent and t independent variable. In order to solve this differential equation, we have to use method of solution of linear differential equation of first order and first degree. Similarly, the differential equation of R C circuit. A circuit involving resistance R and capacitance C along with an electromotive force E of t in a series shown in the figure. This is the R C circuit along with the electromotive force E of t in a series. Here we are nothing but voltage drop across resistance which is equal to R into I. V suffix C indicates voltage drop across capacitance C which is equal to Q by C. Again by Kirchhoff's voltage law sum of voltage drop across resistance R and capacitance C is equal to total electromotive force applied to the circuit. Therefore, the differential equation of this R C circuit is R I plus Q by C equal to E of t. But I indicates the current which is equal to dQ by dt. Therefore, R into dQ by dt plus Q by C equal to E of t. So, this is the required differential equation of R C circuit in a series. This is also belongs to linear differential equation of first order and first degree with Q as a dependent variable and t is an independent variable. So, in order to solve this type of the differential equation we have to use method of linear differential equation. Now, let us pause the video for a while and write the answer to the question. Question is write an integrating factor of the differential equation dI by dt plus 6t into I equal to tQ. Come back I hope you have written answer to this question. Here I will go into explain the solution. Question is write an integrating factor of the differential equation dI by dt plus 6t into I equal to tQ. When we observe this differential equation it is a linear differential equation of first order and first degree with I dependent and t independent variable that is of the type dy by dx plus py equal to Q. When we compare these two equations the value of p here 6 into t and value of Q is tQ. So, these are the functions of variable t. Now, the integrating factor of this differential equation can be obtained by the formula I of equal to e to power integration of p with respect to t which is equal to e to power integration of value of p is 60 into dt is equal to e to power 6 as it is and integration of t is t square by 2. Now, 6 by 2 is 3 therefore, integrating factor is equal to e to power 3 into t square this is the required answer. Now, let us consider example one a circuit having resistance of 20 ohm and an inductance of 10 Henry is connected to 100 volt supply. Determine the value of current after 2 seconds when I equal to 0 at t equal to 0. From the given problem we are given R resistance equal to 20 ohm inductance L equal to 10 Henry and electromotive force E of t equal to 100 volt. When we observe this problem it is a R L circuit. So, that we start with the differential equation of R L circuit that is L into di by dt plus R into i equal to E of t. In this differential equation substitute value of inductance L R and E of t L is 10 Henry therefore, 10 into di by dt plus R is 20 ohm therefore, 20 i is equal to E of t is 100 volt. So, that 10 di plus 20 i equal to 100. Now, dividing 10 on both the side we get di by dt plus 2 into i equal to 10 denote this equation as 1. This is the differential equation of given problem it is a linear differential equation of the type dy by dx plus py equal to q here p is 2 and q is equal to 10. Now, to solve this differential equation first of all we have to find it is integrating factor that is i f equal to e to power integration of p with respect to t which is equal to e to the power integration of value of p is 2 and into dt which is equal to e to the power 2 as it is and integration of dt is t therefore, integrating factor is equal to e to the power 2 into t. Now, general solution of the differential equation 1 is i into integrating factor is equal to integration of integrating factor into capital Q dt plus k where k is constant of integration. In this substitute value of integrating factor and value of capital Q therefore, i into e raise to 2 t is equal to integration of e to the power 2 t it is integrating factor then into value of q is a 10 into dt plus k therefore, i into e to the power 2 t is equal to 10 is taken out as a constant and integration of e raise to 2 t is e raise to 2 t upon 2 plus constant k therefore, i into e raise to 2 t equal to 10 by 2 is 5 into e raise to 2 t plus k. Now, dividing e raise to 2 t on both the side we get current i equal to 5 plus k into e raise to minus 2 t denote this equation by 2. Now, in order to find the value of constant k we have to use the q 1 condition q 1 condition is i equal to 0 at t equal to 0 put in equation 2 we get 0 equal to 5 plus k into e raise to 0 which implies 0 equal to 5 plus k gives k equal to minus 5 put this value of k in equation 2 we get current i equal to 5 plus minus 5 into e raise to minus 2 t. Therefore, i equal to 5 is common taken outside and in bracket 1 minus e raise to minus 2 t denote this equation by 3 this is the required current at any time t flowing in the circuit, but according to the problem we have to find current after 2 second. Therefore, we have to put value of t equal to 2 second in equation 3 we get i is equal to 5 in bracket 1 minus e raise to minus 2 into 2 which is equal to 5 in bracket 1 minus e raise to minus 4. Therefore, i equal to 5 in bracket 1 minus value of e raise to minus 4 is 0.018315639 bracket close. Now, after simplification we get value of i equal to 4.9084 ampere this is the required current flowing in the circuit after 2 second. To prepare this video lecture I refer these 3 as a references. Thank you.